Q148 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
The total number of degrees of freedom associated with $2 \mathrm{~cm}^3$ of Nitrogen gas at normal temperature and pressure is:
[Given Avogadro number as ' N ' ]
- A$\dfrac{N}{4480}$
- B$\dfrac{N}{44800}$
- C$\dfrac{N}{22400}$
- Correct answerD$\dfrac{N}{2240}$
Solution
At normal temperature and pressure (NTP), the molar volume of an ideal gas is $22400 \text{ cm}^3$.
The number of moles $n$ in $2 \text{ cm}^3$ of Nitrogen gas is given by $n = \dfrac{V}{V_{m}} = \dfrac{2}{22400} = \dfrac{1}{11200}$ moles.
Nitrogen ($N_2$) is a diatomic gas. The degrees of freedom $f$ for a diatomic molecule at room temperature is $5$ (3 translational and 2 rotational).
The total number of molecules $N_{total}$ is $n \times N$, where $N$ is the Avogadro number. Thus, $N_{total} = \dfrac{N}{11200}$.
The total degrees of freedom is given by $f \times N_{total} = 5 \times \dfrac{N}{11200} = \dfrac{N}{2240}$.
Answer: $\dfrac{N}{2240}$
The number of moles $n$ in $2 \text{ cm}^3$ of Nitrogen gas is given by $n = \dfrac{V}{V_{m}} = \dfrac{2}{22400} = \dfrac{1}{11200}$ moles.
Nitrogen ($N_2$) is a diatomic gas. The degrees of freedom $f$ for a diatomic molecule at room temperature is $5$ (3 translational and 2 rotational).
The total number of molecules $N_{total}$ is $n \times N$, where $N$ is the Avogadro number. Thus, $N_{total} = \dfrac{N}{11200}$.
The total degrees of freedom is given by $f \times N_{total} = 5 \times \dfrac{N}{11200} = \dfrac{N}{2240}$.
Answer: $\dfrac{N}{2240}$
Hence, the correct answer is option: 'D'.